Q. 73 - Calculus

Question

For each function f and interval [a, b] given in Exercises 68–73, find a real number c ∈ (a, b) such that f(c) is the average value of f on [a, b]. Then use a graph of f to verify that your answer is reasonable.

$f\left(x\right)={(x+1)}^2, [a, b] = [-1, 2]$

Step-by-Step Solution

Answer

The real number is $1,$ and the average value is $3$ and it is verified from the graph.

Step 1. Given Information.

The given function and the interval is $f\left(x\right)={(x+1)}^2, [a, b] = [-1, 2].$

Step 2. Finding a real number c ∈ (a, b).

To find a real number c ∈ (a, b) such that f(c) is the average value of f on [a, b]. We will use the average value formula: $\frac{1}{b-a}{\int }_a^{b}f\left(x\right) dx.$

$$\begin{gathered} \frac{1}{b-a}{\int }_a^{b}f\left(x\right) dx \\ =\frac{1}{2-(-1)}{\int }_{-1}^2{(x+1)}^2 dx=\frac{1}{3}{\int }_{-1}^2{x}^2+2x+1 dx \\ =\frac{1}{3}{\left[\frac{x^3}{3}+\frac{2x^2}{2}+x\right]}_{-1}^2 \\ =\frac{1}{3}\left[\frac{2^3}{3}+\frac{2{\left(2\right)}^2}{2}+2-\left(\frac{{\left(-1\right)}^3}{3}+\frac{2{\left(-1\right)}^2}{2}+\left(-1\right)\right)\right] \\ =\frac{1}{3}\left[\frac{8}{3}+6+\frac{1}{3}-1+1\right] \\ =\frac{1}{3}\left(\frac{27}{3}\right) \\ =3 \end{gathered}$$

Thus, the average value is $3.$

Now,

$$\begin{gathered} f\left(c\right)=3 \\ {\left(c+1\right)}^2=3 \\ {c}^2+2c+1-3=0 \\ {c}^2+2c-2=0 \\ \left(c-1\right)\left(c+2\right)=0 \\ c-1=0 and c+2=0 \\ c=1 and c=-2 \end{gathered}$$

Therefore, the real value is $1, 1\in \left[-1,2\right].$

Step 3. Verification.

By using the graphing utility, the graph of f  is 

From the graph, we can depict that $f\left(c\right)=3.$ Thus, the real number is $1, 1\in \left[-1,2\right].$ Hence, the answer is right.